Question

If $f:(2,3)\to (0,1)$ is defined by $f\left(x\right)=x-\left[x\right],$ then $f^{(-1)}\left(x\right)$ is equal to ,where [.] represents greatest interger function.

Answer 1

Finding the value of ${\mathit{f}}^{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{:}$ :

Given that $f:(2,3)\to (0,1)$ and $f\left(x\right)=x-\left[x\right],$

We know

$\left[x\right]=2,x\in \left(2,3\right)$

$f\left(x\right)=x-2$ which is a line with positive slope.

Therefore one -one always and for onto

Range

$$\begin{gathered} x\in \left(2,3\right) \\ x-2\in \left(0,1\right) \\ f\left(x\right)\in \left(0,1\right) \end{gathered}$$

Therefore function is onto and Bijective.

$$\begin{gathered} Lety=x-2 \\ \Rightarrow x=y+2 \\ \Rightarrow f^{-1}\left(x\right)=x+2 \end{gathered}$$

$\therefore f^{-1}\left(x\right)=x+2$

Hence, the value of ${\mathit{f}}^{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ is $\mathbf{}{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}$